Lemma 1.

ΘLsubscriptnormal-ΘL\Theta_{L} defined above is a congruence relation on AAA.

Proof.

Since LLL is an ideal∅∈LL\varnothing\in L. Therefore, (a,a)∈ΘLaasubscriptnormal-ΘL(a,a)\in\Theta_{L}, since {k∈I∣a⁢(k)≠a⁢(k)}=∅conditional-setkIakak\{k\in I\mid a(k)\neq a(k)\}=\varnothing. Clearly, ΘLsubscriptnormal-ΘL\Theta_{L} is symmetric. For transitivity, suppose (a,b,(b,c)∈ΘLfragmentsnormal-(anormal-,bnormal-,fragmentsnormal-(bnormal-,cnormal-)subscriptnormal-ΘL(a,b,(b,c)\in\Theta_{L}. If a⁢(k)≠c⁢(k)akcka(k)\neq c(k) for some k∈IkIk\in I, then either a⁢(k)≠b⁢(k)akbka(k)\neq b(k) or b⁢(k)≠c⁢(k)bkckb(k)\neq c(k) (a contrapositiveargument). So

Since LLL is an ideal, supp⁡(a,c)∈LsuppacL\operatorname{supp}(a,c)\in L, so (a,c)∈ΘLacsubscriptnormal-ΘL(a,c)\in\Theta_{L}, and ΘLsubscriptnormal-ΘL\Theta_{L} is an equivalence relation on AAA.

Next, let ωω\omega be an nnn-aryoperator on AAA and aj≡bj(modΘL)subscriptajannotatedsubscriptbjpmodsubscriptnormal-ΘLa_{j}\equiv b_{j}\;\;(\mathop{{\rm mod}}\Theta_{L}), where j=1,…,nj1normal-…nj=1,\ldots,n. We want to show that ω⁢(a1,…,an)≡ω⁢(b1,…,bn)(modΘL)ωsubscripta1normal-…subscriptanannotatedωsubscriptb1normal-…subscriptbnpmodsubscriptnormal-ΘL\omega(a_{1},\ldots,a_{n})\equiv\omega(b_{1},\ldots,b_{n})\;\;(\mathop{{\rm mod%
}}\Theta_{L}). Let ωisubscriptωi\omega_{i} be the associated nnn-ary operators on AisubscriptAiA_{i}. If ω⁢(a1,…,an)⁢(k)≠ω⁢(b1,…,bn)⁢(k)ωsubscripta1normal-…subscriptankωsubscriptb1normal-…subscriptbnk\omega(a_{1},\ldots,a_{n})(k)\neq\omega(b_{1},\ldots,b_{n})(k), then ωk⁢(a1⁢(k),…,an⁢(k))≠ωk⁢(b1⁢(k),…,bn⁢(k))subscriptωksubscripta1knormal-…subscriptanksubscriptωksubscriptb1knormal-…subscriptbnk\omega_{k}(a_{1}(k),\ldots,a_{n}(k))\neq\omega_{k}(b_{1}(k),\ldots,b_{n}(k)), which implies that aj⁢(k)≠bj⁢(k)subscriptajksubscriptbjka_{j}(k)\neq b_{j}(k) for some j=1,…,nj1normal-…nj=1,\ldots,n. This implies that

Since LLL is an ideal, and each supp⁡(aj,bj)∈LsuppsubscriptajsubscriptbjL\operatorname{supp}(a_{j},b_{j})\in L, we have that supp⁡(ω⁢(a1,…,an),ω⁢(b1,…,bn))∈Lsuppωsubscripta1normal-…subscriptanωsubscriptb1normal-…subscriptbnL\operatorname{supp}(\omega(a_{1},\ldots,a_{n}),\omega(b_{1},\ldots,b_{n}))\in L as well, this means that ω⁢(a1,…,an)≡ω⁢(b1,…,bn)(modΘL)ωsubscripta1normal-…subscriptanannotatedωsubscriptb1normal-…subscriptbnpmodsubscriptnormal-ΘL\omega(a_{1},\ldots,a_{n})\equiv\omega(b_{1},\ldots,b_{n})\;\;(\mathop{{\rm mod%
}}\Theta_{L}).
∎

Definition. Let A=∏{Ai∣i∈I}Aproductconditional-setsubscriptAiiIA=\prod\{A_{i}\mid i\in I\}, LLL be a Boolean ideal of P⁢(I)PIP(I) and ΘLsubscriptnormal-ΘL\Theta_{L} be defined as above. The quotient algebraA/ΘLAsubscriptnormal-ΘLA/\Theta_{L} is called the LLL-reduced direct product of AisubscriptAiA_{i}. The LLL-reduced direct product of AisubscriptAiA_{i} is denoted by ∏L{Ai∣i∈I}subscriptproductLconditional-setsubscriptAiiI\prod_{L}\{A_{i}\mid i\in I\}. Given any element a∈AaAa\in A, its image in the reduced direct product ∏L{Ai∣i∈I}subscriptproductLconditional-setsubscriptAiiI\prod_{L}\{A_{i}\mid i\in I\} is given by [a]⁢ΘLasubscriptnormal-ΘL[a]\Theta_{L}, or [a]a[a] for short.

Example. Let A=A1×⋯×AnAsubscriptA1normal-⋯subscriptAnA=A_{1}\times\cdots\times A_{n}, and let LLL be the principal idealgenerated by111. Then L={∅,{1}}L1L=\{\varnothing,\{1\}\}. The congruenceΘLsubscriptnormal-ΘL\Theta_{L} is given by (a1,…,an)≡(b1,…,bn)(modΘL)subscripta1normal-…subscriptanannotatedsubscriptb1normal-…subscriptbnpmodsubscriptnormal-ΘL(a_{1},\ldots,a_{n})\equiv(b_{1},\ldots,b_{n})\;\;(\mathop{{\rm mod}}\Theta_{L})iff{i∣ai≠bi}=∅conditional-setisubscriptaisubscriptbi\{i\mid a_{i}\neq b_{i}\}=\varnothing or {1}1\{1\}. This implies that ai=bisubscriptaisubscriptbia_{i}=b_{i} for all i=2,…,ni2normal-…ni=2,\ldots,n. In other words, ΘLsubscriptnormal-ΘL\Theta_{L} is isomorphic to the direct product of A2×⋯×AnsubscriptA2normal-⋯subscriptAnA_{2}\times\cdots\times A_{n}. Therefore, the LLL-reduced direct product of AisubscriptAiA_{i} is isomorphic to A1subscriptA1A_{1}.

For a∈A=∏{Ai∣i∈I}aAproductconditional-setsubscriptAiiIa\in A=\prod\{A_{i}\mid i\in I\}, write a=(ai)i∈IasubscriptsubscriptaiiIa=(a_{i})_{{i\in I}}. It is not hard to see that the mapf:∏P⁢(J){Ai∣i∈I}→∏{Ai∣i∈I-J}normal-:fnormal-→subscriptproductPJconditional-setsubscriptAiiIproductconditional-setsubscriptAiiIJf:\prod_{{P(J)}}\{A_{i}\mid i\in I\}\to\prod\{A_{i}\mid i\in I-J\} given by f⁢([a])=(ai)i∈I-JfasubscriptsubscriptaiiIJf([a])=(a_{i})_{{i\in I-J}} is the required isomorphism.

Remark. The definition of a reduced direct product in terms of a Boolean ideal can be equivalently stated in terms of a Boolean filterFFF. All there is to do is to replace supp⁡(a,b)suppab\operatorname{supp}(a,b) by its complement: supp(a,b)c:={k∈I∣a(k)=b(k)}fragmentssuppsuperscriptfragmentsnormal-(anormal-,bnormal-)cassignfragmentsnormal-{kInormal-∣afragmentsnormal-(knormal-)bfragmentsnormal-(knormal-)normal-}\operatorname{supp}(a,b)^{c}:=\{k\in I\mid a(k)=b(k)\}. The congruence relation is now ΘF′subscriptnormal-ΘsuperscriptFnormal-′\Theta_{{F^{{\prime}}}}, where F′={I-J∣J∈F}superscriptFnormal-′conditional-setIJJFF^{{\prime}}=\{I-J\mid J\in F\} is the ideal complement of FFF. When FFF is prime, the F′superscriptFnormal-′F^{{\prime}}-reduced direct product is called a prime product, or an ultraproduct, since any prime filter is also called an ultrafilter. Ultraproducts can be more generally defined over arbitrary structures.

Mathematics Subject Classification

Comments

Is it possible using these model-theoretic tools to show that if I is an infinite set and \prod_I K_i is a product of fields and if M is a nonprincipal maximal ideal of this ring (thus corresponding to a nonprincipal ultrafilter) then \prod_I K_i/M has cardinality greater than that of K_i for any i in I?

Not my area but I saw talk recently by Kazhdan where he talked of this type of result. He mumbled something about the implicit use of the continuum hypothesis -- so probably not entirely provable but otherwise the techinques are still model theoretic. That would be my only contribution to your question. I'll check back soon and see if someone smarter answers with a better insight.

Take any product of fields R:=\prod_I K_i, with I infinite (it is proven btw that the factor fields of this ring are uncountable and non-archimidean, but more is not known.. as far as I know), take a nonprinicipal maximail ideal of this ring (which corresponds to a nonprincipal ultrafilter of I). Let K=R/M be the factor field wrt M,
then, R':=Kx\prod_I K_i has a factor field from a non-principal maximal ideal (which corresponds to a non-principal ultrafilter of I) which is K, K is also the factor field of the principal maximal ideal (0,1)R' .. (R' can be considered as a tuple here). So the two "ultraproducts" have the same cardinality.

When and where did you saw the talk of Kazhdan? What was the title of his talk? I would be interested in it. Although, this is also not my area.

Another counterexample is as follows: let I be infinite, and each K_i be Z_2, then R = \prod_I K_i is a Boolean ring. Let I be the ideal generated by functions f_i whose ith value is 1 and 0 everywhere else. Then the constant function whose value is 1 is not in I. Extend I to a prime ideal M excluding f. This is possible because R is Boolean. By the same reason, M is maximal. Furthermore, M is non-principal, for if g\in R generates M, then g is 0 at some j. But then f_j\in I does not lie in M, a contradiction. Finally, R/M has order 2 because M is a maximal ideal in a Boolean ring.

You are telling me that you have an ultraproduct from a nonprincipal ultrafilter by which has order 2, which is the same as the order of the fields in your product, in fact in your example you are saying R/M=Z_2 (although I doubt this!)... Why has R/M order 2? You argue that this is because R is a boolean ring.. but how does this show R/M has order 2 if M is nonprincipal. In fact I do believe in this special case R/M has order >2 for M nonprincipal. By "order" you surely mean the cardinality, right? not the characterization of the field R/M. I know this to be a counterexample too, maybe you meant order >2 :) .. in fact I believe one can say more for this special case..

I believe I have read something in these lines:
If R/M is an ultrapower instead of an ultraproduct, i.e. the K_i's in R are all isomorphic to say K, then R/M for M non-principal has cardinality strictly greater than K

> You are telling me that you have an ultraproduct from a
> nonprincipal ultrafilter by which has order 2, which is the
> same as the order of the fields in your product, in fact in
> your example you are saying R/M=Z_2 (although I doubt
> this!)... Why has R/M order 2? You argue that this is
> because R is a boolean ring.. but how does this show R/M has
> order 2 if M is nonprincipal. In fact I do believe in this
> special case R/M has order >2 for M nonprincipal. By "order"
> you surely mean the cardinality, right? not the
> characterization of the field R/M. I know this to be a
> counterexample too, maybe you meant order >2 :) .. in fact I
> believe one can say more for this special case..

Fact: if R is Boolean, and M any maximal ideal in R (principal or not), then R/M is isomorphic to Z_2:

if x is an element of R not in M, then R=(x,M), so that 1=x+m for some m in M. As a result, 1+x=1-x=m is in M. This shows that x+M is either 0+M or 1+M, depending on whether x is in M or not. In other words, R/M is isomorphic to {0,1}.

For more info, please see, for example, Halmo's book "Lectures on Boolean Algebras" (1974, Springer)... great book, I wish they reprint this!

> I believe I have read something in these lines:
> If R/M is an ultrapower instead of an ultraproduct, i.e. the
> K_i's in R are all isomorphic to say K, then R/M for M
> non-principal has cardinality strictly greater than K

Do you remember what your source is?

I found, in "Models and Ultraproducts, an Introduction" by Bell and Slomson (1969, North-Holland), the following:

Suppose I is countably infinite, and {s_i : i in I} is a set such that each s_i is a finite cardinal. Let F be an ultrafilter on I, and P the ultraproduct \prod s_i/F. Then

1. if the set {i in I : s_i=n} is in F for some finite cardinal n, then the card(P)=n,

My reference was a Fundamentae Mathematica paper written by Dana Scott, Morel and Frayne. The paper is very old (also in the 60's) and I had a bad copy of it.. it basically states the fact you just wrote in the end of your post. except that they used "{i in I : s_i=n}\not\in F" in the second part of what you stated, and I read "\in" instead of "\not\in" as the print was very bad.

Yes thanks, having R as Boolean does makes it easier as all other elements are orthogonal to each other.

There is something that is bothering me the whole time. You do know that (x,M)=Rx+M, how can you make sure that x+m =1, I can see that xr+m=1 for some r in R and m in M, but not x+m=1. Or is x+M already R?

Although from the fact you mentioned, R/M must indeed be Z_2 as
{i in I : |Z_2|=2} = I contains the ultrafilter making our maximal ideal.

Another way: if R is any commutative ring with 1, and M is any maximal ideal, then R/M is simple. Any commutative simple ring must be a field. So, if R is Boolean, M is maximal in R, then R/M is simple, and a field. But R has characteristic 2, R/M must be Z_2.